$I(A)$ and $I(B)$ ideal lattices, then $F(J) = \downarrow \psi(J)$ and $G(U)=\downarrow \phi(U)$ is a connection of Galois between $I(A)$ and $I(B)$.

Let $A$ and $B$ be bounded lattices, $\mathcal{I}(A)$ and $\mathcal{I}(B)$ the ideal lattices of $A$ and $B$ and let $(\phi, \psi)$ be a Galois connection between $A$ and $B$: show that $\forall J \in \mathcal{I}(B)\,\,$ we have$\,\,\, \phi^{-1}(J) = \downarrow \psi(J)$, that is the lower order ideal generated by $\psi(J)$. Then prove that $(F,G)$ where $F(J) = \downarrow \psi(J)$ and $G(I)=\downarrow \phi(I)$ is a connection of Galois between $\mathcal{I}(A)$ and $\mathcal{I}(B)$.

Thus $\forall x\in A$ and $\forall y \in \phi^{-1}(J)$ if $x \leq y\,\,\,$we have $\,\,\,\phi(x) \leq \phi (y) \in J \,\,$ that implies $\phi(x)\in J$, that is $x \in \phi^{-1}(J)$, which shows that it is a down-set. $\,\,\,\,$Now I want to see that $\forall x, y \in \phi^{-1}(J) \Rightarrow x \vee y \in \phi^{-1}(J)$, that proves that $\phi^{-1}(J)$ is actually an ideal of $A$: $\,\,\,$if $x, y \in \phi^{-1}(J)$, then $\phi(x), \phi(y) \in J$, hence $\phi(x)\vee \phi(y) \in J\,\,\,$[...]

How to complete it? An any idea to show that $\phi^{-1}(J) = \downarrow \psi(J)$?

Then how to prove that $\forall \,\, I \in \mathcal{I}(A), \forall \,\,J \in \mathcal{I}(B)\,\,\,$ we have $\,\,\,\, \downarrow \phi(I) \subseteq J \iff I \subseteq \downarrow \psi(J)$?

Solution 1:

We have $a\le \psi(b)\iff \phi(a)\le b$ for any $a\in A,\ b\in B$.

Then, $a\in\phi^{-1}(J)\iff \phi(a)\in J\iff \exists b\in J:\phi(a)\le b\iff\exists b\in J:a\le\psi(b)\iff a\in\,\downarrow \psi(J)$.

For the second claim, assume $\downarrow\phi(I)\subseteq J$, that is, if $b\le\phi(a)$ for some $a\in I$ then $b\in J$.
Now let an $a\in I$ be given, and set $b:=\phi(a)$. By the condition, $b\in J$, and we also have $\phi(a)\le b$ so $a\le \psi(b)$, which implies $a\in\, \downarrow\psi(J)$.

The reverse implication can be done similarly.